How many moles of potassium sulfate would you need to prepare 1200 grams of 5.0% sulfate (m/m) solution?
Calculate the volume of solution of 3.6 NH4CI needed to prepare 350.0ml of 1.24 M NH4CI.
How many grams of ammonium chloride are in the solution you created in the above problem.
Well, that's a lot of chemistry questions! But don't worry, I'm here to make it more entertaining for you.
For the first question, to prepare a 5% sulfate solution, you'll need to use your mole-calculating skills. I'll spare you the long explanation, but the answer is it takes 14.4 moles of potassium sulfate.
Now let's move on to the NH4CI dilemma. We need to calculate the volume of a 3.6 solution to prepare 350.0 ml of a 1.24 M NH4CI solution. Let's just say it's a balancing act between the two.
As for the final question, since we already know the concentration and volume of the solution, 1.24 M NH4CI in 350.0 ml, we can calculate the grams using the good old formula: mass (g) = concentration (M) x volume (L) x molar mass (g/mol).
While I hope I brought some laughter to your quest for answers, don't forget to double-check my calculations!
To determine the number of moles of potassium sulfate needed to prepare 1200 grams of a 5.0% sulfate solution, we first need to calculate the mass of sulfate in the solution.
Step 1: Calculate the mass of sulfate in the solution.
Mass percent is defined as the mass of the solute divided by the mass of the solution, multiplied by 100.
Mass percent = (Mass of sulfate / Mass of solution) × 100
Let x be the mass of sulfate in the solution (in grams).
5.0% = (x / 1200 grams) × 100
5.0% = 0.05
0.05 = (x / 1200 grams)
x = 0.05 × 1200 grams
x = 60 grams
Step 2: Calculate the number of moles of potassium sulfate needed.
The molar mass of potassium sulfate (K2SO4) is 174.2592 grams/mole.
Number of moles = Mass of sulfate / Molar mass of potassium sulfate
Number of moles = 60 grams / 174.2592 grams/mole
Number of moles ≈ 0.344 moles (rounded to three decimal places)
Therefore, you would need approximately 0.344 moles of potassium sulfate to prepare 1200 grams of a 5.0% sulfate (m/m) solution.
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To calculate the volume of a 3.6 M NH4Cl solution needed to prepare 350.0 ml of a 1.24 M NH4Cl solution, we can use the dilution formula.
The dilution formula is given by:
C1V1 = C2V2
Where:
C1 = Initial concentration (3.6 M)
V1 = Initial volume (unknown)
C2 = Final concentration (1.24 M)
V2 = Final volume (350.0 ml)
Step 1: Rearrange the dilution formula to solve for V1.
V1 = (C2V2) / C1
V1 = (1.24 M * 350.0 ml) / 3.6 M
V1 ≈ 120.83 ml (rounded to two decimal places)
Therefore, you would need approximately 120.83 ml of a 3.6 M NH4Cl solution to prepare 350.0 ml of a 1.24 M NH4Cl solution.
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To calculate the number of grams of ammonium chloride in the solution created in the previous problem, we can use the formula:
Mass = Concentration * Volume
Step 1: Calculate the mass of ammonium chloride.
Mass = 1.24 M * 350.0 ml
Mass ≈ 434 grams (rounded to three decimal places)
Therefore, in the solution created in the previous problem, there are approximately 434 grams of ammonium chloride.
To calculate the number of moles of potassium sulfate needed to prepare a solution:
Step 1: Determine the mass of sulfate in the desired solution.
Start by calculating the mass of sulfate in the solution using the given percentage by mass. The information given states that the solution is 5.0% sulfate (m/m) and that the total mass of the solution is 1200 grams.
Mass of sulfate = (Percentage of sulfate / 100) x Mass of solution
Mass of sulfate = (5.0 / 100) x 1200 grams
Mass of sulfate = 60 grams
Step 2: Calculate the number of moles of sulfate.
To calculate the number of moles, you need to know the molar mass of potassium sulfate. The molar mass of potassium sulfate (K2SO4) is obtained by adding the atomic masses of its constituent elements (2 x atomic mass of potassium + atomic mass of sulfur + 4 x atomic mass of oxygen).
Molar mass of K2SO4 = (2 x atomic mass of potassium) + atomic mass of sulfur + (4 x atomic mass of oxygen)
Molar mass of K2SO4 = (2 x 39.10 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
Molar mass of K2SO4 = 174.26 g/mol
Now, use the equation:
Moles = Mass / Molar mass
Moles of sulfate = 60 grams / 174.26 g/mol
Step 3: Calculate the moles of potassium sulfate.
Since the empirical formula of potassium sulfate (K2SO4) contains one mole of sulfate, then the number of moles of potassium sulfate is also equal to 60 grams / 174.26 g/mol.
So, you would need approximately 0.344 moles of potassium sulfate to prepare 1200 grams of a 5.0% sulfate solution.
To calculate the volume of solution of 3.6 M NH4Cl needed to prepare 350.0 mL of 1.24 M NH4Cl:
Step 1: Use the equation:
M1V1 = M2V2
where,
M1 = initial concentration of the solution
V1 = initial volume of the solution
M2 = final concentration of the solution
V2 = final volume of the solution
Given that:
M1 = 3.6 M (initial concentration)
V1 = unknown (initial volume)
M2 = 1.24 M (final concentration)
V2 = 350.0 mL (final volume)
Applying the equation:
(3.6 M) x (V1 mL) = (1.24 M) x (350.0 mL)
Step 2: Solve for V1 (initial volume):
V1 mL = (1.24 M x 350.0 mL) / 3.6 M
V1 mL ≈ 121.11 mL
Therefore, you would need approximately 121.11 mL of 3.6 M NH4Cl solution to prepare 350.0 mL of 1.24 M NH4Cl solution.
To calculate the grams of ammonium chloride in the solution created:
Step 1: Calculate the number of moles of ammonium chloride.
Use the formula:
Moles = Molarity x Volume (in liters)
Given that the concentration (molarity) of the solution is 1.24 M, and the volume is 350.0 mL (which is 0.3500 L):
Moles of NH4Cl = 1.24 M x 0.3500 L
Step 2: Calculate the grams of ammonium chloride using the molar mass of NH4Cl.
The molar mass of NH4Cl is obtained by adding the atomic masses of its constituent elements (atomic mass of nitrogen + 4 x atomic mass of hydrogen + atomic mass of chlorine).
Molar mass of NH4Cl = atomic mass of nitrogen + (4 x atomic mass of hydrogen) + atomic mass of chlorine
Molar mass of NH4Cl = 14.01 g/mol + (4 x 1.01 g/mol) + 35.45 g/mol
Molar mass of NH4Cl = 53.49 g/mol
Now, use the equation:
Grams = Moles x Molar mass
Grams of NH4Cl = (1.24 M x 0.3500 L) x 53.49 g/mol
Step 3: Calculate the grams of ammonium chloride.
Grams of NH4Cl ≈ 23.81 grams
Therefore, there are approximately 23.81 grams of ammonium chloride in the solution created in the above problem.
5% SO4 w/w = 5g SO4/100 g solution.
(x g solute/1200) = 0.05
Solve for x g SO4^2- and convert to g K2SO4, then to mols K2SO4.
2. 3.6 what?
3. See 2.